Question

Question #2997a

Answer 1

`"2.5 g sucrose"`

Explanation:

Assuming that you're dealing with percent concentration by mass, `"% m/m"`, you can determine the mass of sucrose present in `"1.0 kg"` of that solution by using its concentration as a conversion factor.

This is essentially what a solution's concentration does -- it helps you go from, in this case, mass of solution to mass of solute or vice versa.

A `0.25%` by mass solution will contain `"0.25 g"` of sucrose, your solute, for every `"100 g"` of solution. So, if every `"100 g"` of this solution will contain `"0.25 g"` of sucrose, it follows that `"1.0 kg"` will contain

`1.0 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * overbrace("0.25 g sucrose"/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 0.25% m/m")) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.5 g sucrose")color(white)(a/a)|)))`